Hamilton's Equations

Hamilton's Equations

As discussed before, Hamiltonian is defined as $$H\equiv \sum_\alpha p_\alpha \dot{q}_\alpha -L$$ The total differential of the Lagrangian is $$dL=\sum_\alpha\left(\frac{\partial L}{\partial q_\alpha}dq_\alpha+\frac{\partial L}{\partial \dot{q}_\alpha}d\dot{q}_\alpha\right)+\frac{\partial L}{\partial t}dt.$$ The total differential of the Hamiltonian is $$dH = \sum_\alpha(\dot{q}_\alpha dp_\alpha + p_\alpha d\dot{q}_\alpha)-dL.$$ So, $$\begin{align} dH &= \sum_\alpha\left(\dot{q}_\alpha dp_\alpha + p_\alpha d\dot{q}_\alpha - \frac{\partial L}{\partial q_\alpha}dq_\alpha - \frac{\partial L}{\partial \dot{q}_\alpha}d\dot{q}_\alpha\right)-\frac{\partial L}{\partial t}dt \\ &= \sum_\alpha\left[\dot{q}_\alpha dp_\alpha + \left(p_\alpha - \frac{\partial L}{\partial \dot{q}_\alpha}\right)\right]-\frac{\partial L}{\partial t}dt. \end{align}$$ Define the generalized momentum as $$p_\alpha \equiv \frac{\partial L}{\partial \dot{q}_\alpha}.$$ Note that the generalized momentum may not be the same as momentum if the potential energy depends on the generalized velocity, i.e. the rate of change of the generalized coordinates $\dot{q}$.
As the generalized momentum is defined as $$p_\alpha \equiv \frac{\partial L}{\partial \dot{q}_\alpha} \implies p_\alpha - \frac{\partial L}{\partial \dot{q}_\alpha}=0,$$ the total derivative of the Hamiltonian becomes $$dH = \sum_\alpha\left(\dot{q}_\alpha dp_\alpha - \frac{\partial L}{\partial q_\alpha}dq_\alpha\right)-\frac{\partial L}{\partial t}dt.$$ The total derivative of the Hamiltonian can also be expressed as $$dH = \sum_\alpha\left(\frac{\partial H}{\partial q_\alpha}dq_\alpha+\frac{\partial H}{\partial p_\alpha}dp_\alpha\right)+\frac{\partial H}{\partial t}dt.$$ Compare the terms, we have $$\bbox[5px,border:2px solid #666] { \dot{q}_\alpha = \frac{\partial H}{\partial p_\alpha}\\ \dot{p}_\alpha = \frac{\partial L}{\partial q_\alpha}=-\frac{\partial H}{\partial q_\alpha}\\ \frac{\partial H}{\partial t} = -\frac{\partial L}{\partial t} }$$ These equations are called Hamilton's equations, or canonical equations.

Example

Find the equations of motion of a spherical pendulum of mass $m$ and length $l$.

Solution: The kinetic energy is $$T=\frac{1}{2}ml^2\dot{\theta}^2+\frac{1}{2}ml^2\sin^2\theta\dot{\phi}^2.$$ The potential energy is $$U=-mgl\cos\theta.$$ The Lagrangian is $$L = \frac{1}{2}ml^2\dot{\theta}^2+\frac{1}{2}ml^2\sin^2\theta\dot{\phi}^2 + mgl\cos\theta$$ The generalized momenta are $$\begin{align} p_{\theta} &= \frac{\partial L}{\partial \dot{\theta}} = ml^2\dot{\theta} \\ p_{\phi} &= \frac{\partial L}{\partial \dot{\phi}} = ml^2\sin^2\theta\dot{\phi}. \end{align}$$ The Hamiltonian is $$\begin{align} H &= T + U \\ &= \frac{1}{2}ml^2\frac{p_{\theta}^2}{m^2l^4} + \frac{1}{2}\frac{ml^2\sin^2\theta p^2_{\phi}}{m^2l^4\sin^4\theta} - mgl\cos\theta \\ &= \frac{p^2_{\theta}}{2ml^2} + \frac{p^2_{\phi}}{2ml^2\sin^2\theta} - mgl\cos\theta. \end{align}$$ The Hamilton's equations are $$\begin{align} \dot{\theta} &= \frac{\partial H}{\partial p_{\theta}} = \frac{p_{\theta}}{ml^2} \\ \dot{\phi} &= \frac{\partial H}{\partial p_{\phi}} = \frac{1}{ml^2\sin^2\theta}p_{\phi} \\ \dot{p}_{\theta} &= - \frac{\partial H}{\partial \theta} = \frac{p^2_{\phi}\cos\theta}{ml^2\sin^3\theta} - mgl\sin\theta\\ \dot{p}_{\phi} &= - \frac{\partial H}{\partial \phi} = 0 \end{align}$$

Checkpoint (Marion, Thornton Q7.29)

A pendulum consists of a mass $m$ and a massless spring with unextended length $b$ and force constant $k$. The pendulum's point of support is rising vertically at a constant acceleration $a$. Find the Hamilton's equations.

The coordinates are $$\begin{align} x &= l\sin\theta \\ y &=\frac{1}{2}at^2 - l\cos\theta. \end{align}$$ So, the velocity components are $$\begin{align} \dot{x} &= \dot{l}\sin\theta + l\dot{\theta}\cos\theta \\ \dot{y} &= at - \dot{l} \cos\theta + l\dot{\theta}\sin\theta. \end{align}$$ The kinetic energy is $$T = \frac{1}{2}m(\dot{x}^2+\dot{y}^2).$$ The potential energy is $$U = mgy + \frac{1}{2}k(l-b)^2.$$ The Lagrangian is $$\begin{align} L &= T - U \\ &= \frac{1}{2}m[\dot{l}^2+l^2\dot{\theta}^2+a^2t^2+2at(l\dot{\theta}\sin\theta-\dot{l}\cos\theta)]\\ &+mg\left(l\cos\theta-\frac{1}{2}at^2\right) - \frac{k}{2}(l-b)^2. \end{align}$$ The generalized momenta are $$\begin{align} p_l &= \frac{\partial L}{\partial \dot{l}}=m\dot{l} - mat\cos\theta\\ p_{\theta} &= \frac{\partial L}{\partial \dot{\theta}} = ml^2\dot{\theta} + matl\sin\theta. \end{align}$$ Invert the generalized momenta, we have $$\begin{align} \dot{l} &= \frac{p_l}{m} + at\cos\theta \\ \dot{\theta} &= \frac{p_{\theta}}{ml^2} - \frac{at\sin\theta}{l}. \end{align}$$ The Hamiltonian is $$\begin{align} H &= \sum{p_i\dot{q}_i}-L \\ &= p_l\dot{l}+p_{\theta}\dot{\theta}-L\\ &= \frac{p^2_l}{2m}+\frac{p^2_\theta}{2ml^2}-\frac{at}{l}p_{\theta}\sin\theta+atp_l\cos\theta\\ &+\frac{1}{2}k(l-b)^2+\frac{1}{2}mgat^2-mgl\cos\theta. \end{align}$$ The Hamilton's equations are $$\begin{align} \dot{p}_{l} + \frac{at}{l^2}p_{\theta}\sin\theta+k(l-b)-mg\cos\theta+\frac{p^2_{\theta}}{ml^3} &=0\\ \dot{p}_{\theta}-\frac{at}{l}p_{\theta}\cos\theta-atp_l\sin\theta+mgl\sin\theta &= 0 \end{align}$$

Time derivative of Hamiltonian

From the Hamilton's equations, we know that $$\frac{\partial H}{\partial t} = -\frac{\partial L}{\partial t}.$$ From the definition of the Hamiltonian, $$\frac{dH}{dt} = -\frac{\partial L}{\partial t}.$$ That means $$\frac{\partial H}{\partial t} = \frac{dH}{dt}.$$ Analyze the total time derivative of Hamiltonian, $$\begin{align} \frac{dH}{dt}&=\sum_\alpha\left(\frac{\partial H}{\partial q_\alpha}\dot{q}_\alpha + \frac{\partial H}{\partial p_\alpha}\right)+\frac{\partial H}{\partial t}\\ &= \sum_\alpha\left[\frac{\partial H}{\partial q_\alpha}\left(\frac{\partial H}{\partial p_\alpha}\right)+\frac{\partial H}{\partial p_\alpha}\left(-\frac{\partial H}{\partial q_\alpha}\right)\right]+\frac{\partial H}{\partial t}\\ &= \frac{\partial H}{\partial t}. \end{align}$$

Legendre Tranformation

Suppose a function $f(x,y)$ is known. Define a new function with new variable $x,p$, $$g(x,y,p)\equiv py-f(x,y)$$ The total differential of $g(x,y,p)$ is $$\begin{align} dg &= pdy + ydp - \frac{\partial f}{\partial x}dx - \frac{\partial f}{\partial y}dy\\ &= \left(p-\frac{\partial f}{\partial y}\right)dy + ydp - \frac{\partial f}{\partial x}dx. \end{align}$$ Define $$p = p(x,y) \equiv \frac{\partial f}{\partial y}$$ such that $$dg = ydp - \frac{\partial f}{\partial x}dx.$$ As the total differential of $g$ is $$dg = \frac{\partial g}{\partial x}dx + \frac{\partial g}{\partial p}dp,$$ we have $$\frac{\partial g}{\partial p}=y$$ and $$\frac{\partial g}{\partial x}=-\frac{\partial f}{\partial x}.$$ We can use these kind of transformation to get back $f$ if $g$ is given or $g$ if $f$ is give. That is to say, we can use Legendre transformation to obtain Lagrangian if Hamiltonian is given, or obtain Hamiltonian if Lagrangian is given.

Example

Given a function $$L(x,y)\equiv \frac{1}{2}my^2-V(x).$$ Apply Legendre transformation to generate a new function. Show that apply the Legendre transformation again will obtain the original function.

Solution: $$p\equiv \frac{\partial L}{\partial y}=my$$ Then, $$y=\frac{p}{m}.$$ A new function is defined as $$H(x,p)\equiv py-L=\frac{p^2}{2m}+V(x).$$ To get back to original function, define $$y\equiv\frac{\partial H}{\partial p}=\frac{p}{m}.$$ Then, $$p=my.$$ A new function is defined as $$\begin{align} L_{\text{new}}(x,y)&\equiv yp - H\\ &= \frac{1}{2}my^2-V(x)\\ &= L(x,y), \end{align}$$ which is equivalent to the original function $L(x,y)$.